Fall 2009
Physics 105, sections 1 and 2
Exam 2
Colton 2-3669
RED
barcode here
Please write your CID here _______
No time limit. No notes. No books. Testing Center calculators only.
Constants:
g = 9.8 m/s2 → but you may use 10
m/s2 in nearly all cases
-11
G = 6.67  10 Nm2/kg2
kB = 1.381  10-23 J/K
NA = 6.022  1023
R = kB∙NA = 8.314 J/mol∙K
 = 5.67  10-8 W/m2∙K4
Mass of Sun = 1.991  1030 kg
Mass of Earth = 5.98  1024 kg
Conversion factors
1 inch = 2.54 cm
1 m3 = 1000 L
Radius of Earth = 6.38  106 m
Radius of Earth’s orbit = 1.496  1011 m
Density of water: 1000 kg/m3
Density of air: 1.29 kg/m3
Specific heat of aluminum: 900 J/kgC
Young’s modulus of steel: 20  1010 N/m2
Linear exp. coeff. of copper: 17  10-6 /C
Linear exp. coeff. of steel: 11  10-6 /C
sin(30) = 0.5
cos(30)  0.866
tan(30)  0.577
  3.14
Specific heat of water: 4186 J/kgC
Specific heat of ice: 2090 J/kgC
Specific heat of steam: 2010 J/kgC
1 atm = 1.013  105 Pa = 14.7 psi
Latent heat of melting (water): 33  105 J/kg
Latent heat of boiling (water): 2.26  106 J/kg
Thermal conduct. of aluminum: 238 J/smC
TK  TC  273.15
9
T F  TC  32
5
Other equations
b  b2  4ac
2a
2
Surface area of sphere = 4 r
x
Volume of sphere =  4 3  r
vave 
3
vi  v f
L  r p  rp  rp sin 
stress = F/A; strain = L L
Y = stress/strain
P  P0  gh
2
v  vo  at
x  xo  vot 
Idisk = (1/2) mR2
Irod (center) = (1/12) mL2
Irod (end) = (1/3) mL2
1 2
at
2
VFR = A1v1 = A2v2
1
1
2
2
P1  v1  gy1  P2  v 2  gy 2
2
2
v f 2  vo 2  2ax
L  L0 T
w = mg, PEg = mgy
F = -kx, PEs = ½ kx2
f  k N (or f   s N )
V  V0 T ;   3
P  F/ / v  Fv cos
F t  p
Elastic: (v1 - v2)bef = (v2 - v1)after
arc length: s  r
v  r
atan  r
ac = v2/r
Fg 
GMm
GMm
, PE g  
2
r
r
Ipt mass = mR2
Isphere = (2/5) mR2
Ihoop = mR2
KE ave 
1
3
2
mvave  k B T
2
2
Q  mcT ; Q  mL
T  T1
Q
 kA 2
T
L
P  e AT 4
|Won gas| = area under P-V curve
= P V (constant pressure)
3
3
U  Nk BT  nRT (monatomic
2
2
ideal gas)
Qh = |Wnet| + Qc
e
Wnet
Qadded
emax  1 
 1
Qc
Qh
Tc
Th
k
, T  2
m
g
, T  2

L

T
v
v

L
g
,  m L
Y ,
v

m
k
B

vair = 343 m/s at 20 C
 I 
-12
2
  10 log   I0 = 10 W/m
I
 0
v  v0
v  vS
v
fn  n
n  1,2,3,...
2L
v
fn  n
n  1,3,5,...
4L
f ' f
Did you write down your CID at the top of the page? _______ If not, you may not get this test booklet back.
1
Instructions:
 Record your answers for problems 1-28 on the bubble sheet.
 You may write on this exam booklet, and are strongly encouraged to do so.
 I strongly encourage you to mark your answers in this test booklet. You will get this test booklet back (but only if
you write your CID at the top of the first page).
 In all problems, ignore friction, air resistance, and the mass of all springs, pulleys, ropes, cables, strings etc.,
unless specifically stated otherwise.
 Use g = 9.8 m/s2 only if there are “9.8” numbers in the answer choices; otherwise use g = 10 m/s2.
 Problems 1-28 will be scaled to be worth 92 total points; problems 29 and 30 are worth 4 points each.
Problem 1. The figure shows a velocity vs time graph of a car moving
along a road, and positive means “to the right”. According to the graph,
what is the car doing from 0 to 5 seconds?
a. moving to the left, slowing down, and stopping
b. moving to the left and speeding up
c. moving to the right, slowing down, and stopping
d. moving to the right and speeding up
e. first moving right, then moving left
f. first moving left, then moving right
velocity
20
10
time
1
2
3
4
5
10
20
1. Negative velocity = “moving to left”; positive velocity = “moving to right”. Choice F
Problem 2. A man holding a rifle at height y = 0 tries to hit the side of a large barn 90 m away. The rifle is shot at 24.7º
above the horizontal. The initial velocity of the bullet is 200 m/s. What is the y-position in meters of the hole the bullet
makes in the side of the barn? Use sin(24.7) = 0.4, cos(24.7) = 0.9. Hint: first find the time it will take.
a. 33.75 m
b. 34.75
c. 35.75
d. 36.75
e. 37.75
f. 38.75
g. 39.75 m
2. First divide v0 into components: v0x = v0cos = 200*0.9 = 180 m/s. v0y = v0sin = 200*0.4 = 80 m/s.
Then, x-direction gives time: x = v0xt → 90 = 180 t → t = 0.5 seconds
Finally, y-direction gives final height: y = v0yt – ½ gt2 → y = 80*0.5 – ½*10*(0.5)2
y = 40 – 5*0.25 = 20 – 1.25 = 38.75 m. Choice F
Problem 3. From the perspective of Newton’s Second Law, what force on a car causes it to accelerate forward when the
gas pedal is pushed?
a. The car pushing backward on the road
b. The car pushing forward on the road
c. The road pushing backward on the car
d. The road pushing forward on the car
3. This must be a force on the car, which is pushing forward. Choice D.
Problem 4. You slide a block of ice down a 45ramp. There is no friction between the ice and the ramp. How will the
acceleration of the ice block down the ramp compare to the standard acceleration of gravity, 9.8 m/s2?
a. It will be equal to 9.8 m/s2.
b. It will be greater than 9.8 m/s2.
c. It will be less than 9.8 m/s2.
4. As discussed in class, it must be between 0 and 9.8 m/s2 (the values when  = 0 and  = 90). Choice C.
2
Problem 5. A boat moves through the water with two forces acting on it. One is a 2200 N forward push by the water on
the propeller, and the other is a 1500 N resistive force due to the water around the bow. What is the acceleration of the
1000 kg boat?
a. Less than 0.53 m/s
b. 0.53 – 0.63
c. 0.63 – 0.73
d. 0.73 – 0.83
e. 0.83 – 0.93
f. More than 0.93 m/s
F = ma
2200 – 1500 = 1000a
a = 700/1000 = 0.7 m/s2. Choice C.
Problem 6. You need to carry a suitcase up a flight of stairs. In which case will you do the most work on it against
gravity?
a. You carry the suitcase up quickly.
b. You carry the suitcase up slowly.
c. Both cases involve the same amount of work.
6. The work against gravity will be equal to F//x, which will also equal the change in potential energy:
mgh. Doesn’t matter has fast you carry the suitcase. Choice C.
Problem 7. A woman with a parachute is falling at constant speed. How does the magnitude of the force of gravity
pulling down on the woman compare to the magnitude of the force of the parachute pulling up on her?
a. Gravity is grater
b. Parachute is greater
c. They are the same
7. This is just like the monkey sliding down a pole problem. If the speed is constant, there is no
acceleration and the forces must add up to zero. Therefore the forces from gravity and the parachute are
equal (and opposite direction, obviously). Choice C.
Problem 8. On a frozen pond, an 8 kg sled has an initial speed of v0 = 10 m/s. The coefficient of kinetic friction between
sled and ice is k = 0.2. Find the distance the sled moves before coming to rest.
a. Less than 9 m
b. 9 – 15
c. 15 – 21
d. 21 – 27
e. 27 – 33
f. 33 – 39
g. More than 39 m
8. Ebef + Wnet = Eaft
½ mv2 – fx = ½ mvf2
Note that f = N, and from the y-direction, N = mg. Therefore,
2
½*8*10 – 0.2*8*10*x = 0
Divide by 8, and you get
½*100 – 2x = 0
x = 50/2 = 25 m. Choice D
3
Problem 9. The amount of potential energy possessed by an object that has been lifted up is equal to:
a. the distance the object is lifted
b. the force used to lift the object
c. the object’s acceleration due to gravity
d. the weight of the object
e. the work done in lifting the object
9. Potential energy accounts for the work done by conservative forces. Choice E
Problem 10. Jen throws three balls from the top of a building towards a large flat parking lot: one straight up, one
straight down and one horizontally, all with the same initial speed. Which ball will have the highest speed just before it
hits the ground? (As usual, ignore air resistance.)
a. The one thrown horizontally
b. The one thrown straight down
c. The one thrown straight up
d. All three will all have the same speed
10. They all have the same initial kinetic and potential energy. They all have the same final potential
energy. Therefore if there was no work done by non-conservative forces (i.e., air resistance), the final
kinetic energy must be the same. Choice D.
Problem 11. A box slides down a ramp. The work done on the box by the normal force is:
a. positive
b. negative
c. zero
11. W = F//x. Since the normal force is perpendicular to x, there is no parallel component and the work
done is zero.
Problem 12. The parachute on a race car of mass 600 kg opens when the car is traveling at 50 m/s. What average force
(magnitude) must be supplied by the parachute to stop the car in a time of 20 seconds?
a. Less than 1550 N
b. 1550 – 1650
c. 1650 – 1750
d. 1750 – 1850
e. 1850 – 1950
f. 1950 – 2050
g. More than 2050 m
12. First, get acceleration through kinematics:
vf = v0 + at
0 = 50 + a*20
a = -5/2 m/s2
Now use N2 to get force:
F = ma
Fparachute = ma
Fparachute = 600*(-5/2) = -1500 N, 1500 N magnitude
4
Problem 13. A skier of mass 71 kg is pulled up a 15 frictionless slope by a motor-driven cable at a constant speed of 3
m/s. How many horsepower must a motor have to perform this task? (1 hp = 745.7 W)
713
3
a.
hp
e.

9.8 sin 15  745.7
719.8 sin 15 745.7

b.


71
9.8 sin 15  3745.7


f.
sin 15 3

c.
d.

719.83
sin 15 745.7
71sin 15 3 hp


g.
719.8745.7
9.8745 .7
719.8sin 15  3
745 .7
13. Start with: P = Fv, since both force and velocity are constant.
From FBD, Ftension - mgsin = 0, F = mgsin
So P = (mgsin)v = 719.8sin153 watts
Convert to horsepower by multiplying by (1 hp)/(745.7 W)
719.8 sin 15  3 . Choice D
P=
745 .7


Problem 14. You are a passenger in a car and not wearing your seat belt. Without increasing or decreasing its speed, the
car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is the correct analysis of the
situation according to Newton’s laws?
a. Before and after the collision with the door, there is a rightward force pushing you into the door.
b. Starting at the time of collision with the door, the door exerts a leftward force on you.
c. Both of the above
d. Neither of the above
14. Ignoring the fictitious forces which seem to be present in the accelerating reference frame of inside the
car, it’s clear that your own inertia wants to keep going straight, which is why you collide with the door
when the car goes left. After you collide, though, the door exerts a force on you (otherwise you would fly
out of the car). Choice B.
Problem 15. In the “racing balls” demo, two balls were raced on different tracks. The closer track had a section that was
lower for part of the race. Which ball won the race, and why?
a. The closer ball, because it was going a greater speed during the lower section.
b. The closer ball, because it had a greater change of potential energy from start to finish.
c. The farther away ball, because it was going a greater speed during the lower section.
d. The farther away ball, because it had a greater change of potential energy from start to finish.
e. The balls tied, because they both ended with the same kinetic energy.
15. Choice A describes the situation exactly. (They both end up with the same vf, though.)
Problem 16. Newton’s 3rd Law says that forces come in pairs. A book lying on a table feels a force of gravity and a
normal force. What is the Newton’s 3rd Law partner force with the force of gravity on the book?
a. the gravitational force of the book pulling up on the earth
b. the normal force pushing up on the book
c. the pressure of the book pushing down on the table
d. none of the above
16. The partner force of F1-2 is F2-1. Thus, the partner of Fearth-book is Fbook-earth. That is, the force of the book
pulling up on the earth. Choice A
5
Problem 17. Two cables support a cat burglar of weight W. One cable is at an angle, , as
shown in the figure. Find the tension in the cable connected to the left wall, in terms of W
and .
w
a.
cos
w
b.
sin 
w
c.
tan 
d. w cos 
e. w sin 
T2sin
f. w tan 
T2
17. FBD of junction:

T1
T2cos
w
N2 in y-direction: Fy = may
T2sin - w = 0
T2 = w/sin
N2 in x-direction: Fx = max
T2cos – T1 = 0
(w/sin)*cos


T1 = w/tan. Choice C.
Problem 18. A hanging mass, m1, is attached via a pulley to another mass, m2, which is
resting on a horizontal frictionless table as shown in the figure. Find the acceleration of the
masses in terms of m1, m2, and g.
a. g
b. m1 g
c. m 2 g
m1 g
d.
m1  m2
m2 g
e.
m1  m2
f.
g.
 m1  m2  g
m1
 m1  m2  g
m2
18. N2 for group, in the “bent” x-direction:
Fgroup = mgroupa
m1g = (m1+m2)a
a = m1g/(m1+m2). Choice D
6
m2
m1
Problem 19. Repeat, but with coefficient of friction  between the table and m2. You may take it as given that the
friction force is small enough that m1 still does fall. Note: A free body diagram and an equation will be required for this
situation, in problem 29.
m1 g   m2 g
a.
m1  m2
FBD:
 m2 g
N
b.
T
m1  m2
m2 g   m2 g
c.
m2
f
m1
m1  m2
T
m1 g   m2 g
d.
m1  m2
m1 g
m1 g
m2 g
e.
m1   m2
m1 g   m2 g
f.
m1   m2
m1 g
g.
 m2
19. N2 for group, in the “bent” x-direction:
Fgroup = mgroupa
m1g – f = (m1+m2)a
m1g - N = (m1+m2)a
Get normal force from y-direction: Fm2y = 0 → N = m2g
m1g - m2g = (m1+m2)a
a = (m1g- m2g)/(m1+m2). Choice A
Problem 20. Same situation as the last problem. This time, however, the friction force is strong enough to prevent m1
from falling. Specifically, m1 = 1 kg, m2 = 7 kg,  = 0.5, and both masses are at rest. What is the tension in the line
connecting the two masses?
a. Less than 8 N
b. 8 – 16
c. 16 – 24
d. 24 – 32
e. 32 – 40
f. 40 – 48
g. More than 48 N
20. N2 for m1: Fm1= m1a. Note a = 0, since masses are not moving
T - m1g = 0
T = m1g = 1*10 = 10N. Choice B
7
Problem 21. You want to measure the spring constant of a hanging spring. You add a 0.5 kg mass to the initially
unmoving spring and quickly release the mass so that it starts oscillating up and down with no additional forces on it. At
the lowest point of the first oscillation, the mass has stretched the spring out to a distance of 20 cm below where it
started. What is the spring constant k?
a. Less than 30 N/m
b. 30 – 60
c. 60 – 90
d. 90 – 120
e. 120 – 150
f. More than 150 N/m
21.
Before:
After
h
y=0
Ebef + Wnet = Eaft
mgh + 0 = ½ kh2
k = 2mg/h = 2*0.5*10/0.2 = 50 N/m. Choice B
Problem 22. A 3 gram ping-pong ball starts at rest and is pushed 0.5 meters along an evacuated tube by a constant force
of 30 N from air entering the tube behind it. How long does it take the ball to exit the tube?
a. 0.01 s
b. 0.02 s
c. 0.03 s
d. 0.04 s
e. 0.05 s
f. 0.06 s
g. 0.07 s
F = ma
30 = 0.003a
a = 10000 m/s2
x = x0 + v0xt + ½at2
0.5 = 0 + 0 + 1/2 *10000t2
t=sqrt(1/10000) = 1/100 s. choice A
8
Problem 23. A block (m = 0.4 kg) accelerates down a 35 ramp (measured from the horizontal) at a rate of 0.5 m/s2.
There is friction on the ramp. What is k?
0.5
a.
4 9.8cos37
b.
c.
d.
e.
f.


2  9.8sin 37   0.5
4  9.8  9.8cos37
0.5
9.8cos37   0.5 
2
3
free body diagram
9.8sin 37  0.5
9.8cos37
f
N
9.8 tan 37  0.5
mgsin
mgcos
23. Fy = 0  N = mgcos
f = N = mgcos
mg
Fx = ma  mgsin – f = ma
mgsin – mgcos = ma
Cancel the m’s: gsin – gcos = a
 = (gsin – a)/(gcos). Choice E
Problem 24. Two identical cars, coasting from rest down two hills having the same vertical height—but one hill steeper
than the other—will arrive at the bottom of the hills with the same speed. Yet the acceleration of the car down the steep
hill is greater than the car going down the other hill. What is the best explanation for this seeming contradiction?
a. The acceleration when we neglect friction and air resistance is always 9.8 m/s2
b. The car going down the less-steep hill accelerates for a longer distance.
c. The car going down the steeper hill experiences more nonconservative forces.
d. There is no contradiction: the cars will arrive at the bottom of the hills at the same time.
24. Similar to the “racing balls” question, although they both end up with the same final speed, the time is
different. This is possible because the one experiencing less acceleration, accelerates for a longer distance
(and a longer time). Choice B
9
Problem 25. Tarzan (80 kg) swings on a vine (length 20 m) initially inclined at an
angle of 36.9 from the vertical. He pushes off with an initial speed of 4 m/s. What is
his speed at the bottom of the swing? The picture may help; Tarzan is the ball on the
end of the vine. Note: sin(36.9) = 0.60, cos(36.9) = 0.80, tan(36.9) = 0.75.
a.
88 m/s
b.
92 m/s
c.
d.
96 m/s
100 m/s
e.
104 m/s
f.
108 m/s

L
y
y=0
This distance is Lcos
This distance is h = L - Lcos
= 20 - 20*0.8
= 20 - 16 = 4 m
25. Ebef + Wnet + Eaft
½mv2 + mgh = ½mvf2
Cancel the m’s, multiple both sides by 2, plug in the numbers:
42 + 2*10*4 = vf2
16 + 80 = vf2
vf = sqrt(96) m/s. Choice C
Problem 26. A crate (mass m) is being pushed across a level floor at a constant speed by a

force F at an angle of  below the horizontal as shown in the figure. In terms of F, , m, and
g, find the coefficient of kinetic friction between the crate and the floor. Note: A free body
diagram and some equations will be required for this situation, in Problem 30.
1
a.  
tan 
F
b.  
mg
F
c.  
mg  F sin 
F
d.  
mg  F tan 
F cos
e.  
mg
F cos
f.  
mg  F sin 
free body diagram
N
26. Fy = 0 → N = mg + Fsin
Constant speed: Fx = 0 → Fcos = N
mg + Fsin = Fcos
 = Fcos/mg + Fsin. Choice F
Fcos

f=N
mg
10
Fsin
Problem 27. A block of mass m = 2.7 kg is held without moving on a frictionless
incline of angle of 19 by the horizontal force F, as shown in the figure. Determine
the value of the force F, in Newtons.
a. 2.7  9.8 Newtons
2.7  9.8
b.
cos19
2.7  9.8
c.
sin19
2.7  9.8
d.
tan19
free body diagram
e. 2.7  9.8  cos19
Fcos
f. 2.7  9.8  sin19
N
F
g. 2.7  9.8  tan19 Newtons
27. Fy = 0
Fcos – mgsin = 0
F = mgsin/cos = mgtan. Choice G.
mg
Problem 28. Vern, the famous “pulley ski-jumper” is 100
kg, and is pulled via a pulley by a 300 kg hanging block up
a 50 m incline that has a vertical height change of 20 m; see
the figure. Friction provides a constant backwards force on
Vern of 100 N. How fast will Vern be going when he takes
off from the incline, assuming he misses the giant pulley?
a. Less than 620 m/s
b. Between
620 and
640
c. Between
660
d. Between
640 and
660 and
e. Between
680 and
700
f.
More than
Fcos
mgsin
50 m
20 m
50 m
680
mgcos
Final
positions
700 m/s
28. Ebef + Wnet = Eaft
PEbig mass – Ffriction*x = KEboth masses + PEskier
Note that while the skier is raised 20 m, the big mass falls 50 m. See figure. Also, friction acts over 50 m.
300kg*10m/s2*50m – 100N*50m = ½ (100kg+300kg)vf2 + 100kg*10m/s2*20m
150000 – 5000 – 20000 = 200vf2
vf2 = 125000/200
vf = sqrt(625). Choice B.
Note—if you use g = 9.8 instead of 10 for this one, you get sqrt(612) as the answer. Although I said
specifically to use g = 10 in the exam instructions at the top, I have decided to accept answer
choice A as well as choice B. I had the Testing Center automatically regrade everyone’s exams, so
you should automatically get credit for answer A if that’s what you answered.
11
Write your CID here: ________
Problem 29 (4 pts). (a) Draw a free-body diagram for each mass in Problem 19. Be sure to label all forces.
FBD:
N
T
m2
m1
f
T
m1 g
m2 g
(b) Write down the N2 “blueprint equation” for the two masses taken as a group, and fill in the blueprint with
the given forces. Don’t solve the equation, just take it one or two steps past the blueprint. [If you don’t know
how to use N2 for a group, do two N2 equations for the two masses separately.]
Blueprint:
Fgroup = mgroupa
…or Fm1 = m1a and Fm2 = m2a
Filled in:
m1g – f = (m1+m2)a
m1g - N = (m1+m2)a (1 step further, not needed)
…or m1g – T = m1a and T - f = m2a
…or m1g – T = m1a and T - N = m2a
Problem 30 (4 pts). (a) Draw a free-body diagram of the crate from Problem 26. Be sure to label all forces.
free body diagram
N
Fcos

f=N
mg
Fsin
(b) Based on your FBD, write down and fill in the N2 “blueprint equations” for the x- and the y-directions.
Don’t solve the equations, just take them one or two steps past the blueprint. Be sure to fill in the acceleration(s)
if known.
x-direction
y-direction
Blueprint:
Fx= max
Fy= may
Filled in:
Fcosf  



FcosN (1 step further, not neeed)
12

- mg - Fsin = 0